\((a)\frac { 7 }{ 10 }\)  
\((b)\frac { 7 }{ 40 } \)

\((a)\frac { 2 }{ 5 } \) 
\((b)\frac { 3 }{ 5 } \)

Let E be the event of hitting the boundary.
Then,
\(P(E)=\frac { Number\ of\ times\ the\ batswoman\ hits\ the\ boundary }{ Total\ number\ of\ balls\ she\ plays } \)
 \(=\frac { 6 }{ 30 } =\frac { 1 }{ 5 } =0.2\)
Probability of not hitting the boundary
= 1- Probability of hitting the boundary
= 1- P(E) = 1 - 0.2 = 0.8

Total number of students = 200
(i) Number of students who like statistics = 135
Probability that a student chosen at random likes statistics=\(\frac { 135 }{ 200 } =\frac { 27 }{ 40 } \)
(ii) Number of students who do not like statistics=65
Probability that a student chosen at random does not like it= \(\frac { 65 }{ 200 } =\frac { 13 }{ 40 } \)
Aliter:
 Probability that a student chosen at random does not like statistics
= 1 - probability that a student chosen at random likes statistics
 =\(1-\frac { 27 }{ 40 } =\frac { 13 }{ 40 } \) 

Total number of students born in the year
= 3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4 = 40
Number of students born in August = 6
. Probability that a student of the class was born in August =\(\frac { 6 }{ 40 } =\frac { 3 }{ 20 } \)

\(\frac { 3 }{ 2 } >1\) so, \(\frac { 3 }{ 2 } \) cannot be the experimental probability of an event.

P(E)+P(F)=1

1,2,3,4,5,6

H,T

\(0\le P(E)\le 1\)

(i) P(sum is 7)\(=\frac{75}{500}=\frac{3}{20}\)
(ii) P(sum is more than 11)\(=\frac{70}{500}=\frac{7}{50}\)
(iii) P(sum is less than or equal to 6)\(=\frac{19+30+22+52+55}{500}\)
\(=\frac{178}{500}=\frac{89}{250}\)
(iv) P(sum is between 5 and 10)\(=\frac{52+75+70+53}{50}\)
\(=\frac{250}{500}=\frac{1}{2}\)

Total student = 7 + 7 + 11 + 3 + 5 + 9 + 8 = 50
(i) P(height of student is more that 135 cm)\(=\frac{25}{50}=\frac{1}{2}\)
(ii)P(height of student is at least 145 cm)\(=\frac{17}{50}\)
(iii) p(height of student is less than 130 cm)\(=\frac{14}{50}=\frac{7}{25}\)
(iv) P(height of student is 125 or more but less than 140)\(=\frac{21}{50}\)

Total number of times a die is rolled = 25
Even numbers are 2, 4, 6.
Probability of getting an even number
\(=\frac { 4+6+0 }{ 25 } =\frac { 10 }{ 25 } =\frac { 2 }{ 5 } \)
(b) Multiples of 3 are 3, 6.
Probability of getting a multiple of 3
\(=\frac { 5+0 }{ 25 } =\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
(c) 2, 3, 5 are prime numbers.
Probability of getting a prime number
\(=\frac { 4+5+1 }{ 25 } =\frac { 10 }{ 25 } =\frac { 2 }{ 5 } \)

Total number of days = 250
(i) Number of days on which the weather forecasts were correct = 175
Probability that on a given day it was correct = \(\frac { 175 }{ 250 } =\frac { 7 }{ 10 } \)
(ii) Probability that it was not correct on a given day=\(1-\frac { 7 }{ 10 } =\frac { 3 }{ 10 } \) 

P(A)+P(B)=1
\(\because \) Sum of the probabilities of all the outcomes of an event is 1
\(\Rightarrow \ 0.72+P(B)=1\)
\(\\ \Rightarrow \ P(B)=1-0.72=0.28\)

\((i)\frac { 33 }{ 50 } \) 
\((ii)\frac { 13 }{ 20 } \)

The total number of days for which the record is available = 250
(i) P(the forecast was correct on a given day)
\( =\frac{\text { Number of days when the forecast was correct }}{\text { Total number of days for which the record is available }} \)
\(=\frac{175}{250}=0.7\)
(ii) The number of days when the forecast was not correct = 250 - 175 = 75
So, P(the forecast was not correct on a given day) \(=\frac{75}{250}=0.3\)

0.179, 0.15, 0.157, 0.149, 0.175, 0.19

\(\frac { 1 }{ 2 } \)

No of women = 70 - 15(+20 + 30)
= 5
P(women) = \(\frac{5}{70}=\frac{1}{14}\)